It would be useful to know some context, and whether this is homework, and where the problem comes from and why it is interesting. There is a neat way of looking at this, but it would be a shame to be given a 'magic' solution out of context because you won't then be able to apply the same idea to a different problem.
–
Mark BennetAug 3 '12 at 9:04

6 Answers
6

I enjoyed working this out-cool question man. Here's a straightforward inductive proof:

I'm writing $c$ instead of $n$.

Inductively assume $(c + \sqrt{c^2 - 1})^k = a + b \sqrt{c^2 - 1}$, where $b^2(c^2 - 1) = a^2 - 1$ (the base case is clear). That is, not only can you write it as $t + \sqrt{t^2 - 1}$, but also $t^2 - 1$ is a square times $c^2 - 1$ (I guessed this by working out the first $k$'s for $c =2,3$).

Then $(n+\alpha)(n-\alpha)=1$ and it is easy to see (prove by induction) that $(n+\alpha)^k = t+u\alpha \text{ and } (n-\alpha)^k = t-u\alpha$ with $t$ and $u$ integers so: $$1=(n+\alpha)^k(n-\alpha)^k = (t+u\alpha)(t-u\alpha)= t^2-u^2\alpha^2 $$

First of all, notice that if $n$ is an integer and $a = n + \sqrt{n^2-1}$, then $n = \frac{1}{2}(a+\frac{1}{a})$. Next notice that the direction of implication can be reversed here, i.e. if $\frac{1}{2}(a+\frac{1}{a})$ is an integer then a can be written in the form $n + \sqrt{n^2-1}$ where $n$ is an integer by setting $n = \frac{1}{2}(a+\frac{1}{a})$.

With this in mind, we simply have to prove that if $\frac{1}{2}(a+\frac{1}{a})$ is an integer then so is $\frac{1}{2}(a^k+\frac{1}{a^k})$ for any integer $k$. This can be easily seen by induction on $k$ using the binomial expansion of $(a+\frac{1}{a})^k$ and the symmetry of the binomial coefficients.

The roots of $x+\frac1x=2n$ are $x_n=n+\sqrt{n^2-1}$ and $1/x_n=n-\sqrt{n^2-1}$.

If we can show that
$$
x_n^k+1/x_n^k=2m_{n,k}\tag{1}
$$
(that is, an even integer), then we get the form we want:
$$
x_n^k=m_{n,k}+\sqrt{m_{n,k}^2-1}\tag{2}
$$
Initial Case: for $k=1$, obviously, $x_n^1+1/x_n^1=2n$.

Inductive Case: suppose that $(1)$ holds for all $j<k$, that is, $x_n^j+1/x_n^j=2m_{n,j}$. Then,
$$
\begin{align}
(2n)^k
&=\left(x_n+1/x_n\right)^k\\
&=x_n^k+1/x_n^k+\sum_{j=1}^{k-1}\binom{k}{j}x_n^{k-2j}\\
&=x_n^k+1/x_n^k+\left\{\begin{array}{}
\binom{k}{k/2}+\sum_{j=1}^{k/2-1}\binom{k}{j}(x_n^{k-2j}+1/x_n^{k-2j})&\mbox{if $k$ is even}\\
\sum_{j=1}^{(k-1)/2}\binom{k}{j}(x_n^{k-2j}+1/x_n^{k-2j})&\mbox{if $k$ is odd}
\end{array}\right.\\
&=x_n^k+1/x_n^k+\left\{\begin{array}{}
\binom{k}{k/2}+2\sum_{j=1}^{k/2-1}\binom{k}{j}m_{n,k-2j}&\mbox{if $k$ is even}\\
2\sum_{j=1}^{(k-1)/2}\binom{k}{j}m_{n,k-2j}&\mbox{if $k$ is odd}
\end{array}\right.\tag{3}
\end{align}
$$
When $k$ is even, $\binom{k}{k/2}$ has as many factors of $2$ as there are $1$-bits in the binary representation of $k$ (that is, at least $1$). Therefore, since $(2n)^k$, $\binom{k}{k/2}$ (if $k$ is even), and $2\sum\binom{k}{j}m_{n,k-2j}$ are all even, $(3)$ says that $x_n^k+1/x_n^k$ is even, which validates $(1)$.